Codeforces Round #710 (Div. 3)-E. Restoring the Permutation

  • Post author:
  • Post category:其他



题目链接


A permutation is a sequence of n integers from 1 to n, in which all numbers occur exactly once. For example, [1], [3,5,2,1,4], [1,3,2] are permutations, and [2,3,2], [4,3,1], [0] are not.

Polycarp was presented with a permutation p of numbers from 1 to n. However, when Polycarp came home, he noticed that in his pocket, the permutation p had turned into an array q according to the following rule:

qi=max(p1,p2,…,pi).

Now Polycarp wondered what lexicographically minimal and lexicographically maximal permutations could be presented to him.

An array a of length n is lexicographically smaller than an array b of length n if there is an index i (1≤i≤n) such that the first i−1 elements of arrays a and b are the same, and the i-th element of the array a is less than the i-th element of the array b. For example, the array a=[1,3,2,3] is lexicographically smaller than the array b=[1,3,4,2].

For example, if n=7 and p=[3,2,4,1,7,5,6], then q=[3,3,4,4,7,7,7] and the following permutations could have been as p initially:

[3,1,4,2,7,5,6] (lexicographically minimal permutation);

[3,1,4,2,7,6,5];

[3,2,4,1,7,5,6];

[3,2,4,1,7,6,5] (lexicographically maximum permutation).

For a given array q, find the lexicographically minimal and lexicographically maximal permutations that could have been originally presented to Polycarp.

Input

The first line contains one integer t (1≤t≤104). Then t test cases follow.

The first line of each test case contains one integer n (1≤n≤2⋅105).

The second line of each test case contains n integers q1,q2,…,qn (1≤qi≤n).

It is guaranteed that the array q was obtained by applying the rule from the statement to some permutation p.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105.

Output

For each test case, output two lines:

on the first line output n integers — lexicographically minimal permutation that could have been originally presented to Polycarp;

on the second line print n integers — lexicographically maximal permutation that could have been originally presented to Polycarp;

题意:有一个数组a,长度为n,a的元素是1-n中每一个数组成,现给定b数组,



b

i

b_i







b










i





















等于



m

a

x

(

a

1

,

a

2

,

a

3

.

.

.

a

i

)

max(a_1,a_2,a_3…a_i)






m


a


x


(



a










1


















,





a










2


















,





a










3


















.


.


.



a










i


















)





,求a数组字典序最小的排列和字典序最大的排列

思路:我们可以根据b数组确定一些a数组元素,字典序最小的通过从小到大填补空缺即可,字典序最大的用set二分查找逐步填入就行

#include<bits/stdc++.h>
#define ll long long
using namespace std;
void slove()
{
    int n;cin>>n;
    int a[n+5];
    int maxans[n+5];
    int minans[n+5];
    map<int,int>mp;
    set<int>s;
    vector<int>v;
    memset(maxans,-1,sizeof maxans);
    memset(minans,-1,sizeof minans);
    for(int i=0;i<n;i++)
        cin>>a[i],mp[a[i]]=1;
    for(int i=1;i<=n;i++)
    {
        if(mp[i]==0)
        {
            s.insert(i);
            v.push_back(i);
        }
    }
    //cout<<s.size()<<endl;
   // for(auto p=s.begin();p!=s.end();++p)
     //   cout<<*p<<endl;
    maxans[0]=a[0];
    minans[0]=a[0];
    for(int i=1;i<n;i++)
    {
        if(a[i]!=a[i-1])
        {
            maxans[i]=a[i];
            minans[i]=a[i];
        }
    }
    int f=0;
    for(int i=1;i<n;i++)
    {
        if(maxans[i]==-1)
        {
            maxans[i]=v[f];
            f++;
        }
    }
   // for(int i=0;i<n;i++)
    //    cout<<minans[i]<<" ";
    //cout<<endl;
    for(int i=1;i<n;i++)
    {
        if(minans[i]==-1)
        {
            if(s.size()==0) break;
            auto p=--s.upper_bound(minans[i-1]);
            minans[i]=*p;
            //cout<<*p<<" "<<minans[i]<<endl;
            s.erase(p);
            //for(auto u=s.begin();u!=s.end();++u)
            //    cout<<*u<<" ";
            //    cout<<endl;
        }
    }
    for(int i=0;i<n;i++)
        cout<<maxans[i]<<" ";
    cout<<endl;
    for(int i=0;i<n;i++)
        cout<<minans[i]<<" ";
    cout<<endl;

}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t;cin>>t;
    while(t--)
        slove();
    return 0;
}



版权声明:本文为small_whl原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接和本声明。